Power Rule

General Power Rule

Let $f$ be a differentiable function and $h (x) = (f (x))^{n}$ for $n \in R$ , then

h^{'} (x) = \frac{d}{d x} (f (x))^{n} = n (f (x))^{n - 1} \cdot f^{'} (x)

For now, this proof is only for $n \in \mathbb{N}$.

Note the fact that the following equality is true:

x^{n} - a^{n} = (x - a) \cdot \sum_{i = 1}^{n} x^{n - i} a^{i - 1}

Let $f (x) = x^{n}$ , we have to show that $f^{'} (a) = n a^{n - 1}$ for any $a$ :

\begin{aligned} f^{'} (a) & = lim_{h \to 0} \frac{f (a + h) - f (a)}{h} \\ = lim_{x \to a} \frac{f (x) - f (a)}{x - a} \\ = lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} \end{aligned}

Using the fact stated above, we have

\begin{aligned} = lim_{x \to a} \frac{(x - a) \cdot \sum_{i = 1}^{n} x^{n - i} a^{i - 1}}{x - a} \\ = lim_{x \to a} \sum_{i = 1}^{n} x^{n - i} a^{i - 1} \\ = \sum_{i = 1}^{n} a^{n - i} a^{i - 1} \\ = \sum_{i = 1}^{n} a^{n - 1} \\ = n a^{n - 1} \end{aligned}

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